Question

A fair die is tossed until a number greater than $4$ appear. The probability that an even number of tosses shall be required is:

• A. $\frac{1}{2}$
• B. $\frac{3}{5}$
• C. $\frac{2}{5}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C $\frac{2}{5}$

Probability of required outcome is $1/3$ for 1st time
Probability of required outcome is $2/3\ast 1/3$ for 2nd time
Probability of required outcome is $2/3\ast 2/3\ast 1/3$ for 3rd time
So we want even number of tosses
Therefore
$P=2/3\ast 1/3+2/3\ast 2/3\ast 2/3\ast 1/3+..........$
Hence solving the G.P
$P=2/5$